Question

According to Langmuir adsorption isotherm, the amount of gas adsorbed at very high pressure:

• A. reaches a constant limiting value
• B. goes on increasing with pressure
• C. goes on decreasing with pressure
• D. increases first and decreases later with pressure

Answer 1

The correct option is A reaches a constant limiting value

In Langmuir adsorption isotherm,
$\theta$ : Fraction of surface covered on adsorbent
$1-\theta$ : Free surface are
$P$ : Pressure of gas

By langmuir adsorption isotherm,
$\text{Rate of adsorption}=k_{a}P(1-\Theta )$
$\text{Rate of desorption}=k_d\Theta$
$k_a$ and $k_d$ are rate constants.

At equilibrium condition,
$\text{Rate of adsorption}=\text{Rate of desorption}$
$k_{a}P(1-\Theta )=k_d\Theta$
$\Theta =\frac{KP}{KP+1}$

According to Langmuir, extent of adsorption is proportional to $\Theta$
$\frac{x}{m}\propto \Theta$
$\frac{x}{m}\propto \frac{KP}{1+KP}$

$\frac{x}{m}=K^{\prime }\frac{KP}{1+KP}$

$\frac{x}{m}=\frac{K^{\prime }KP}{1+KP}$
$\frac{x}{m}=\frac{aP}{1+bP}$
where,
$a=K^{\prime }K=K^{\prime }\frac{k_a}{k_d}$
$b=K=\frac{k_a}{k_d}$

$\frac{x}{m}=\frac{aP}{1+bP}$
At high pressure, $bP>>1$
so, $1+bP\equiv bP$
$\frac{x}{m}=\frac{aP}{bP}$

$\frac{x}{m}=\frac{a}{b}$

$\frac{x}{m}$ becomes constant i.e. the surface is fully covered and change in pressure has no effect and no further adsorption takes place​.